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Question
If for non-zero x, af(x) + bf \[\left( \frac{1}{x} \right) = \frac{1}{x} - 5\] , where a ≠ b, then find f(x).
Solution
Given :
\[af\left( x \right) + bf\left( \frac{1}{x} \right) = \frac{1}{x} - 5\] ...(i)
On adding equations (i) and (ii), we get:
\[af\left( x \right) + bf\left( x \right) + bf\left( \frac{1}{x} \right) + af\left( \frac{1}{x} \right) = \frac{1}{x} - 5 + x - 5\]
\[\Rightarrow \left( a + b \right)f\left( x \right) + \left( a + b \right)f\left( \frac{1}{x} \right) = \frac{1}{x} + x - 10\]
\[\Rightarrow f\left( x \right) + f\left( \frac{1}{x} \right) = \frac{1}{\left( a + b \right)}\left[ \frac{1}{x} + x - 10 \right]\] ...(iii)
On subtracting (ii) from (i), we get:
\[af\left( x \right) - bf\left( x \right) + bf\left( \frac{1}{x} \right) - af\left( \frac{1}{x} \right) = \frac{1}{x} - 5 - x + 5\]
\[\Rightarrow \left( a - b \right)f\left( x \right) - f\left( \frac{1}{x} \right)\left( a - b \right) = \frac{1}{x} - x\]
\[\Rightarrow f\left( x \right) - f\left( \frac{1}{x} \right) = \frac{1}{\left( a - b \right)}\left[ \frac{1}{x} - x \right]\] ...(iv)
On adding equations (iii) and (iv), we get:
\[2f\left( x \right) = \frac{1}{a + b}\left[ \frac{1}{x} + x - 10 \right] + \frac{1}{a - b}\left[ \frac{1}{x} - x \right]\]
\[\Rightarrow 2f\left( x \right) = \frac{\left( a - b \right)\left[ \frac{1}{x} + x - 10 \right] + \left( a + b \right)\left[ \frac{1}{x} - x \right]}{\left( a + b \right)\left( a - b \right)}\]
\[\Rightarrow 2f\left( x \right) = \frac{\frac{a}{x} + ax - 10a - \frac{b}{x} - bx + 10b + \frac{a}{x} - ax + \frac{b}{x} - bx}{a^2 - b^2}\]
\[\Rightarrow 2f\left( x \right) = \frac{\frac{2a}{x} - 10a + 10b - 2bx}{a^2 - b^2}\]
\[\Rightarrow f\left( x \right) = \frac{1}{a^2 - b^2} \times \frac{1}{2}\left[ \frac{2a}{x} - 10a + 10b - 2bx \right]\]
\[= \frac{1}{a^2 - b^2}\left[ \frac{a}{x} - 5a + 5b - bx \right]\]
Therefore ,
\[f\left( x \right) = \frac{1}{a^2 - b^2}\left[ \frac{a}{x} - bx - 5a + 5b \right]\]
\[= \frac{1}{a^2 - b^2}\left[ \frac{a}{x} - bx \right] - \frac{5\left( a - b \right)}{a^2 - b^2}\]
\[= \frac{1}{a^2 - b^2}\left[ \frac{a}{x} - bx \right] - \frac{5\left( a - b \right)}{\left( a - b \right)\left( a + b \right)}\]
\[= \frac{1}{a^2 - b^2}\left[ \frac{a}{x} - bx \right] - \frac{5}{\left( a + b \right)}\]
Hence,
\[f\left( x \right) = \frac{1}{a^2 - b^2}\left[ \frac{a}{x} - bx \right] - \frac{5}{\left( a + b \right)}\]
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